我是android学习,我正在使用Json解析并得到如下问题如何解决这个问题?
org.json.JSONException:类型java.lang.String的值SELECT无法转换为JSONObject
这是我的代码
{"status": 1,
"msg": "success",
"errorCode": "",
"errorDesc": "",
"data": [
{
"sdate": "2016-02-05",
"ID": "3366",
"post_title": "Guided Tours of Louis Armstrong House",
"post_date": "2016-01-15 18:30:30",
"description": "\r\n\tEvery Friday\r\n",
"post_content": "\r\n\tEvery Friday\r\n",
"guid": "http://hothousejazz.com/event_detail.php?eid=3366",
"start_date": "2016-02-05",
"end_date": "2016-03-04",
"evo_hide_endtime": "No",
"venue": "Queens",
"location": "34-56 107th Street, Corona, NY",
"geo_location": "34-56 107th Street, Corona, NY",
"latitude": "40.754596",
"longitude": "-73.8613472",
"phone": "718-478-8274",
"stime": "10:00 AM",
"etime": "05:00 PM",
"website": "http://www.louisarmstronghouse.org"
}
这是我的json回复
02-09 01:12:41.399 1979-2009/? W/System.err: org.json.JSONException: Value SELECT of type java.lang.String cannot be converted to JSONObject
这是log
<?xml version="1.0" encoding="utf-8"?>
<layer-list xmlns:android="http://schemas.android.com/apk/res/android" >
<item>
<shape>
<solid android:color="@color/gray" />
</shape>
</item>
<item
android:bottom="1px"
android:left="0px">
<shape android:shape="rectangle" >
<solid android:color="#FFFFFF" />
</shape>
</item>
</layer-list>
答案 0 :(得分:1)
Jsonobject jsonobj=new Jsonobject(response)
Jsonarray jsonarray=new Jsonarray(jsonobj.getString("status"));
for(int i=0;i<jsonarray.length();i++)
{
Here you can fetch the other values inside Jsonarray
}
这是解析json的方法但你的json格式不正确请看看它
答案 1 :(得分:0)
这样的数据
JSONObject jsonObject = new JSONObject(response);
String status = jsonObject.getString("msg");
if (status.equals("success")){
JSONArray cast = jsonObject.getJSONArray("data");
for (int i=0; i<cast.length(); i++){
JSONObject actor2 = cast.getJSONObject(i);
String name =actor2.getString("sdate");
String note = actor2.getString("id");
String point = actor2.getString("post_title");
String has_image = actor2.getString("post_date");
}
试试这个。